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1. DIFFERENTIATION, Let us Study, •, , Derivatives of Composite functions., , •, , Geometrical meaning of Derivative., , •, , Derivatives of Inverse functions, , •, , Logarithmic Differentiation, , •, , Derivatives of Implicit functions., , •, , Derivatives of Parametric functions., , •, , Higher order Derivatives., Let us Recall, , •, , The derivative of f (x) with respect to x, at x = a is given by, , •, , The derivative can also be defined for f (x) at any point x on the open interval as, . If the function is given as y = f (x) then its derivative is written as, ., , •, , For a differentiable function y = f (x) if δx is a small increment in x and the corresponding increment, in y is δy then, , •, , ., , Derivatives of some standard functions., y = f (x), c (Constant), xn, 1, x, 1, xn, , sin x, cos x, tan x, , y = f (x), 0, nxn−1, 1, − 2, x, −, , n, xn+1, , cos x, − sin x, sec2 x, , sec x, , sec x tan x, , cosec x, , − cosec x cot x, , cot x, , − cosec2 x, , ex, , ex, , ax, , a x log a, , log x, , 1, x, , log a x, Table 1.1.1, , 1, , 1, x log a
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Rules of Differentiation :, If u and v are differentiable functions of x such that, (i), , y = u ± v then, , (iii), , y=, , , , (ii), , y = uv then, , u, where v ≠ 0 then, v, , Introduction :, The history of mathematics presents the development of calculus as being accredited to Sir Isaac, Newton (1642-1727) an English physicist and mathematician and Gottfried Wilhelm Leibnitz (16461716) a German physicist and mathematician. The Derivative is one of the fundamental ideas of calculus., It's all about rate of change in a function. We try to find interpretations of these changes in a mathematical, way. The symbol δ will be used to represent the change, for example δx represents a small change in the, variable x and it is read as "change in x" or "increment in x". δy is the corresponding change in y if y is, a function of x., We have already studied the basic concept, derivatives of standard functions and rules of, differentiation in previous standard. This year, in this chapter we are going to study the geometrical, meaning of derivative, derivatives of Composite, Inverse, Logarithmic, Implicit and Parametric functions, and also higher order derivatives. We also add some more rules of differentiation., Let us Learn, 1.1.1 Derivatives of Composite Functions (Function of another function) :, So far we have studied the derivatives of simple functions like sin x, log x, e x etc. But how about, , log sin (x2 + 5) or e tan x etc ? These are known as composite functions. In, the derivatives of sin, this section let us study how to differentiate composite functions., 1.1.2 Theorem : If y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of, x such that the composite function y = f [g (x)] is a differentiable function of x then, , ., , Proof : Given that y = f (u) and u = g (x). We assume that u is not a constant function. Let there be a, small increment in the value of x say δx then δu and δy are the corresponding increments in u and, y respectively., As δx, δu, δy are small increments in x, u and y respectively such that δx ≠ 0, δu ≠ 0 and δy ≠ 0., We have, , ., , Taking the limit as δx → 0 on both sides we get,, , 2
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As δx → 0, we get, δu → 0 (⸪ u is a continuous function of x), , , . . . . . (I), , Since y is a differentiable function of u and u is a differentiable function of x., we have,, and, , , , . . . . . (II), , From (I) and (II), we get, , , . . . . . (III), , The R.H.S. of (III) exists and is finite, implies L.H.S.of (III) also exists and is finite , . Then equation (III) becomes,, , Note:, 1., , The derivative of a composite function can also be expressed as follows. y = f (u) is a differentiable, function of u and u = g (x) is a differentiable function of x such that the composite function, y = f [ g (x)] is defined then, ., , 2., , If y = f (v) is a differentiable function of v and v = g (u) is a differentiable function of u and u = h (x), is a differentiable function of x then, ., , 3., , If y is a differentiable function of u1, ui is a differentiable function of ui+1 for i = 1, 2, ..., n−1 and un, is a differentiable function of x, then, , This rule is also known as Chain rule., , 3
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1.1.3 Derivatives of some standard Composite Functions :, dy, dx, , y, , dy, dx, , n [ f (x)] n−1 ⋅ f '(x), f '(x), , cot [ f (x)], , − cosec2 [ f (x)]⋅ f '(x), , y, [ f (x)] n, , √ f (x), , cosec [ f (x)] − cosec [ f (x)] ⋅ cot [ f (x)] ⋅ f '(x), , 2√ f (x), , 1, [ f (x)]n, , n ⋅ f '(x), , −, , sin [ f (x)], cos [ f (x)], tan [ f (x)], sec [ f (x)], , [ f (x)] n+1, cos [ f (x)]⋅ f '(x), − sin [ f (x)]⋅ f '(x), sec2 [ f (x)]⋅ f '(x), sec [ f (x)] ⋅ tan [ f (x)] ⋅ f '(x), , a f (x), , a f (x) ⋅ log a ⋅ f '(x), , e f (x), , e f (x) ⋅ f '(x), f '(x), , log [ f (x)], , f (x), f '(x), , log a [ f (x)], , f (x) log a, , Table 1.1.2, SOLVED EXAMPLES, Ex. 1 : Differentiate the following w. r. t. x., (i) y =, , √ x2 + 5 (ii), , (iv) log (x5 + 4) , Solution :, , (i) y =, , (v), , y = sin (log x) (iii), 53 cos x − 2 (vi), , y = e tan x, 3, y=, (2x2 − 7)5, , √ x2 + 5, , Method 1 :, , Method 2 :, , Let u = x + 5 then y = √ u , where y is, a differentiable function of u and u is a, differentiable function of x then, 2, , , , We have y =, , √ x2 + 5, , Differentiate w. r. t. x, , . . . . . (I), [Treat x2 + 5 as u in mind and use the formula, of derivative of √ u ], , Now, y = √ u, Differentiate w. r. t. u, and u = x2 + 5, Differentiate w. r. t. x, du d 2, =, (x + 5) = 2x, dx dx, Now, equation (I) becomes,, , 4
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(ii) y = sin (log x), Method 2 :, , Method 1 :, , We have y = sin (log x), , Let u = log x then y = sin u, where y is, a differentiable function of u and u is a, differentiable function of x then, , , Differentiate w. r. t. x, [sin (log x)], , . . . . . (I), , [Treat log x as u in mind and use the formula, of derivative of sin u], , Now, y = sin u, Differentiate w. r. t. u, = cos u and u = log x, Differentiate w. r. t. x, 1, =, x, Now, equation (I) becomes,, , Note : Hence onwards let's use Method 2., (iii) y = e tan x, , (iv) Let y = log (x5 + 4) , Differentiate w. r. t. x, , Differentiate w. r. t. x, , [log (x5 + 4)], , [e tan x], , (v) Let y = 53 cos x − 2, , (vi) Let y =, , 3, , (2x2 − 7)5, Differentiate w. r. t. x, , Differentiate w. r. t. x, , [53 cos x − 2], dy, , = 53 cos x − 2 · log 5 × (3 cos x − 2), dx, dy, = − 3 sin x · 53 cos x − 2 · log 5, dx, , , , , 5
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= (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 ⋅, , (x + cos x) + (x + cos x) 3⋅ 4(x3 + 2x − 3)3 ⋅, , (x3 + 2x − 3), , = (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 (1 − sin x) + (x + cos x) 3 ⋅ 4(x3 + 2x − 3)3 (3x2 + 2), ∴, , dy, = 3(x3 + 2x − 3)4 (x + cos x) 2 (1 − sin x) + 4 (3x2 + 2) (x3 + 2x − 3)3 (x + cos x)3, dx, , (v) y = (1 + cos2 x) 4 × √ x + √tan x, Differentiate w. r. t. x, , , , , , , , , , Ex. 3 : Differentiate the following w. r. t. x., (i) y = log3 (log5 x) (ii), , (iii) , , (iv), , (v) , , (vi), , 7
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Solution :, (i) y = log3 (log5 x), , , = log3, , = log3 (log x) - log3 (log 5), , ∴ y =, , - log3 (log 5), , Differentiate w. r. t. x, , , , , , , [Note that log3(log 5) is constant], , , ∴, (ii), , , , [ ⸪ log e = 1], , ∴, Differentiate w. r. t. x, , , , , ∴, , 8
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(iii) , , ∴ , Differentiate w. r. t. x, , , , , , , , , , ∴ , , (iv), , , , , , ∴, Differentiate w. r. t. x, , 9
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(vi), , (v), , y = a cot x [⸪ alog a f (x) = f (x)], Differentiate w. r. t. x, , , , , (a cot x ), , [⸪ alog a f (x) = f (x)], , , , = sin2 x + cos2 x, ∴, , = a cot x log a ·, , y =1, Differentiate w. r. t. x, , (cot x), , =a cot x log a (− cosec2 x), − cosec2 x· a cot x log a, , Ex. 4, , :, , Solution :, , , If f (x) = √ 7g (x) − 3 , g (3) = 4 and g' (3) = 5, find f ' (3)., Given that : f (x) = √ 7g (x) − 3, Differentiate w. r. t. x, , , ∴, , , For x = 3, we get, , , , =, , 35, 2(5), , =, , 7, 2, , [Since g (3) = 4 and g' (3) = 5], , 10
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(2) Differentiate the following w.r.t. x, (i), , cos (x2 + a2), , (iii), (v), , (xv), , (ii), (iv), , cot3[log (x3)], , (vii) cosec (√cos x), (ix) e3 sin, , 2 x − 2 cos2 x, , (xvi), , (vi) 5sin, , 3x+3, , (viii) log [cos (x3 − 5)], (x) cos2 [log (x2 + 7)], , (xi) tan [cos (sin x)] (xii) sec[tan (x4 + 4)], (xiii) elog [(log x), , − log x2 ], , (xiv) sin √sin √x, , (xv) log[sec (e )], , (xvi) loge2 (log x), , 2, , x2, , (xvii), , (xvii) [ log [log(log x)]], , (xviii), (xix) y = (25)log5 (sec x) − (16)log4 (tan x), (xx), , 2, , (xviii) sin2 x2 − cos2 x2, , (4) A table of values of f, g, f ' and g' is given, , (3) Differentiate the following w.r.t. x, , (x2 + 4x + 1)3 + (x3 − 5x − 2)4, 5, 8, (ii) (1 + 4x) (3 + x − x2), , x, , f (x), , g(x), , f '(x), , g'(x), , 2, , 1, , 6, , −3, , 4, , 4, , 3, , 4, , 5, , −6, , (iii), , 6, , 5, , 2, , −4, , 7, , (i), , (v), , (iv), , (1 + sin2 x)2 (1 + cos2 x)3, , (i), , If r(x) = f [g(x)] find r' (2)., , (vi) √cos x + √cos √x, , (ii), , If R(x) = g[3 + f (x)] find R' (4)., , (vii) log (sec 3x + tan 3x) (viii), , (iii) If s(x) = f [9 − f (x)] find s' (4)., (iv) If S(x) = g [ g(x) ] find S' (6)., , (ix), (x), , (5) Assume that f ' (3) = − 1, g' (2) = 5, g (2) = 3, (xi), , and y = f [g(x)] then, , (xii) log [tan3 x·sin4 x·(x2 + 7) ], 7, , (6) If, , (xiii), , , f (1) = 4, g (1) = 3,, , f ' (1) = 3, g' (1) = 4 find h' (1)., (7) Find the x co-ordinates of all the points on, the curve y = sin 2x − 2 sin x, 0 ≤ x < 2π, , (xiv), , where, , 12, , 0.
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1.2.2 Derivatives of Inverse Functions :, We know that if y = f (x) is a one-one and onto function then x = f, , −1, , ( y) exists. If f, , −1, , ( y) is, , differentiable then we can find its derivative. In this section let us discuss the derivatives of some inverse, functions and the derivatives of inverse trigonometric functions., x+2, Example 1 : Consider f (x) = 2x − 2 then its inverse is f −1 (x) =, . Let g(x) = f −1 (x)., 2, 1, d, d, [ f (x)] = 2 and, [g (x)] = ., If we find the derivatives of these functions we see that, 2, dx, dx, These derivatives are reciprocals of one another., Example 2 : Consider y = f (x) = x2 . Let g = f −1., ∴ g ( y) = x = √ y, ∴ g' ( y) =, Now, , 1, 2√y, , also f ' (x) = 2x, , d, [g ( f (x))] =, dx, , = 1 and g [ f (x)] = x ∴, , At a point ( x, x2 ) on the curve, f ' (x) = 2x and g' ( y) =, , 1, 2√y, , =, , d, d, [g ( f (x))] = (x) = 1, dx, dx, 1, 1, =, ., 2x f '(x), , 1.2.3 Theorem : Suppose y = f (x) is a differentiable function of x on an interval I and y is One-one, onto and, d −1, dx, dy, 1, ≠ 0 on I. Also if f −1( y) is differentiable on f (I ) then, [ f ( y)] =, or, =, dy, dy, dx, f '(x), , where, , dy, ≠ 0., dx, , Proof : Given that y = f (x) and x = f −1 (y) are differentiable functions., Let there be a small increment in the value of x say δx then correspondingly there will be an, increment in the value of y say δy. As δx and δy are increments, δx ≠ 0 and δy ≠ 0., δx δy, We have,, ×, =1, δy δx, δx, δy, ∴, = , where, ≠0, δy, δx, Taking the limit as δx → 0, we get,, , , as δx → 0, δy → 0,, , , , , Since y = f (x) is a differentiable function of x., , 14, , . . . . . (I)
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we have,, , and, , dy, ≠ 0 , dx, , . . . . . (II), , From (I) and (II), we get, , , As, , dy, ≠ 0,, dx, , . . . . . (III), , δx, exists and is finite. ∴ lim, δy→0, δy, , Hence, from (III), , dx, =, dy, , where, , =, , dx, exists and is finite., dy, , dy, ≠0, dx, , An alternative proof using derivatives of composite functions rule., [Identity function], We know that f −1 [ f (x)] = x, Taking derivative on bothe sides we get,, d, d, [ f −1 [ f (x)]] = (x), dx, d dx, i.e. ( f −1)' [ f (x)], [ f (x)] = 1, dx, i.e. ( f −1)' [ f (x)] f ' (x) = 1, 1, ∴ ( f −1)' [ f (x)] =, , . . . . . (I), f ' (x), So, if y = f (x) is a differentiable function of x and x = f −1 ( y) exists and is differentiable then, , ( f −1)' [ f (x)] = ( f −1)' ( y) =, ∴, , (I) becomes, dx, dy, =, where, ≠0, dy, dx, , dx, dy, and f ' (x) =, dy, dx, , SOLVED EXAMPLES, Ex. 1 : Find the derivative of the function y = f (x) using the derivative of the inverse function, x = f −1 ( y) in the following, (i), , (ii), , (iii) y = ln x, , Solution :, (i), We first find the inverse of the function y = f (x), i.e. x in term of y., y3 = x + 4 ∴ x = y3 − 4 ∴ x = f −1 ( y) = y3 − 4, dy, =, =, dx, for x ≠ −4, , , , 15
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(ii), , , , We first find the inverse of the function y = f (x), i.e. x in term of y., 2, y2 = 1 + √ x i.e. √ x = y2 − 1, ∴ x = f −1 ( y) = ( y2 − 1), , , dy, =, dx, , =, , , , , (iii) y = log x, We first find the inverse of the function y = f (x), i.e. x in term of y., y = log x ∴ x = f −1 ( y) = e y, 1, 1, 1, dy, =, = y = ln x = ., =, x, e, e, dx, Ex. 2 : Find the derivative of the inverse of, function y = 2x3 − 6x and calculate its, value at x = −2., Solution : Given : y = 2x3 − 6x, Diff. w. r. t. x we get,, dy, = 6x2 − 6 = 6 (x2 − 1), dx, dx, =, , we have,, dy, ∴, , Ex. 3 : Let f and g be the inverse functions of, each other. The following table lists a, few values of f, g and f ', , 1, 6 ((−2)2 − 1), , =, , 1, 18, , g(x), , f '(x), , −4, , 2, , 1, , 1, 3, , 1, , −4, , −2, , 4, , Solution : In order to find g' (−4), we should first, find an expression for g' (x) for any input, x. Since f and g are inverses we can use, the following identify which holds for, any two diffetentiable inverse functions., 1, g' (x) =, ... [check, how?], f ' [g(x)], , ... [Hint : f [g(x)] = x], , at x = − 2,, , we get, y = 2(−2)3 − 6(−2), , = − 16 + 12 = − 4, , =, , f (x), , find g' (−4)., , dx, 1, =, 2, dy 6 (x − 1), , , , x, , ∴ g' (−4) =, =, , 16, , 1, f '[g (−4)], 1, , f '(1), , =, , 1, 4
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Ex. 4 : Let f (x) = x5 + 2x − 3. Find ( f −1)' (−3)., Solution : Given : f (x) = x5 + 2x − 3, Diff. w. r. t. x we get,, f '(x) = 5x4 + 2, , , Note that y = −3 corresponds to x = 0., 1, ∴ ( f −1)' (−3) =, f ' (0), 1, 1, =, =, 5(0) + 2 2, 1.2.4 Derivatives of Standard Inverse trigononmetric Functions :, We observe that inverse trigonometric functions are multi-valued functions and because of this,, their derivatives depend on which branch of the function we are dealing with. We are not restricted to, use these branches all the time. While solving the problems it is customary to select the branch of the, inverse trigonometric function which is applicable to the kind of problem we are solving. We have to, pay more attention towards the domain and range., dy, π, π, 1, ≤ y ≤ then prove that, =, , |x| < 1., dx √1 − x2, 2, 2, π, π, Proof : Given that y = sin−1 x, −1 ≤ x ≤ 1, − ≤ y ≤, 2, 2, ∴ x = sin y , . . . . (I), Differentiate w. r. t. y, dx d, = (sin y), dy dy, dx, = cos y = ± √cos2 y = ± √1 − sin2 y, dy, dx, = ± √1 − x2, . . . . [⸪ sin y = x], ∴, dy, π, π, , But cos y is positive since y lies in 1st or 4th quadrant as − ≤ y ≤, 2, 2, dx, 2, ∴, = √1 − x, dy, dy, , We have, =, dx, 1., , If y = sin−1 x, −1 ≤ x ≤ 1, −, , dy, 1, =, , |x| < 1, dx √1 − x2, dy, 1, If y = cos−1 x, −1 ≤ x ≤ 1, 0 ≤ y ≤ π then prove that, =−, ., dx, √1 − x2, [As home work for students to prove.], , ∴, 2., , 17
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3., , If y = cot−1 x, x ∈ R, 0 < y < π then, , dy, 1, =−, ., dx, 1 + x2, , Proof : Given that y = cot−1 x, x ∈ R, 0 < y < π, ∴, , x = cot y , . . . . (I), , Differentiate w. r. t. y, dx d, = (cot y), dy dy, dx, = − cosec2 y = − (1 + cot2 y), , dy, dx, ∴, = − (1 + x2) , . . . . [⸪ cot y = x], dy, dy, =, , We have, dx, , , ∴, , 1, dy, =, , dx − (1 + x2), , dy, 1, =−, dx, 1 + x2, , ∴, , dy, π, π, 1, < y < then, =, . [left as home work for students to prove.], dx 1 + x2, 2, 2, dy, π, 1, =, if x > 1, 5. If y = sec−1 x, such that |x| ≥ 1 and 0 ≤ y ≤ π, y ≠ then, dx x√ x2 − 1, 2, dy, 1, =−, if x < − 1, , dx, x√ x2 − 1, π, Proof : Given that y = sec−1 x, |x| ≥ 1 and 0 ≤ y ≤ π, y ≠, 2, ∴ x = sec y , . . . . (I), Differentiate w. r. t. y, dx d, = (sec y), dy dy, dx, = sec y · tan y, dy, dx, = ± sec y ·√tan2 y, ∴, dy, = ± sec y ·√sec2 y − 1, Fig. 1.2.2, dx, ∴, = ± x √ x2 − 1, . . . . . [⸪ sec y = x], dy, We use the sign ± because for y in 1st and 2nd quadrant. sec y · tan y > 0., Hence we choose x √ x2 − 1 if x > 1 and − x √ x2 − 1 if x < − 1, 4., , If y = tan−1 x, x ∈ R, −, , In 1st quadrant both sec y and tan y are positive., In 2nd quadrant both sec y and tan y are negative., ∴ sec y · tan y is positive in both first and second quadrant., , 18
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Also, for x > 0, x √ x2 − 1 > 0, and for x < 0, − x √ x2 − 1 > 0, dx, = x √ x2 − 1 , , when x > 0, |x| > 1, dy, dy, dx, dy, dx, , = − x √ x2 − 1 , , 1, =, , x√ x2 − 1, 1, =−, x√ x2 − 1, , when x < 0, |x| > 1, , i.e. x > 1, i.e. x < − 1, , if x > 1, if x < − 1, , Note 1 : A function is increasing if its derivative is positive and is decreasing if its derivative is, negative., Note 2 : The derivative of sec−1 x is always positive because the graph of sec−1 x is always increasing., dy, π, π, 1, ≤ y ≤ , y ≠ 0 then, =−, if, dx, 2, 2, x√ x2 − 1, dy, 1, if, =, dx x√ x2 − 1, [ Left as home work for students to prove ], , 6., , If y = − cosec x, such that |x| ≥ 1 and −, , Note 3 : The derivative of cosec−1 x is always negative because the graph of cosec, decreasing., , x>1, x<−1, , −1, , x is always, , 1.2.5 Derivatives of Standard Inverse trigonometric Functions :, dy, dx, , y, , sin −1 x, , cos −1 x, , tan −1 x, , cot −1 x, , 1, √1 − x2, −, , Conditions, , , |x| < 1, , 1, √1 − x2, , , |x| < 1, , 1, 1 + x2, −, , 1, 1 + x2, , −1 ≤ x ≤ 1, π, π, − ≤y≤, 2, 2, −1 ≤ x ≤ 1, 0≤y≤π, x∈R, π, π, − <y<, 2, 2, x∈R, , dy, dx, , Conditions, , 1, for x > 1, x√ x2 − 1, 1, −, for x < − 1, x√ x2 − 1, , |x| ≥ 1, 0≤y≤π, π, y≠, 2, , y, , sec −1 x, , cosec −1 x, , 0<y<π, Table 1.2.1, , 19, , −, , 1, , for x > 1, , x√ x2 − 1, 1, for x < − 1, x√ x2 − 1, , |x| ≥ 1, π, π, − ≤y≤, 2, 2, y≠0
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1.2.6 Derivatives of Standard Inverse trigonometric Composite Functions :, dy, , y, sin −1 [ f (x)], , cos −1 [ f (x)], , tan, , −1, , dx, f ' (x), , √1 − [ f (x)], , 2, , −, , f ' (x), , √1 − [ f (x)]2, , , | f (x)| < 1, , cot −1 [ f (x)], , , | f (x)| < 1, , sec, , −, , dx, f ' (x), 1 + [ f (x)]2, , f ' (x), , f ' (x), , [ f (x)], , dy, , y, , −1, , [ f (x)], , cosec −1 [ f (x)], , 2, , 1 + [ f (x)], , f (x) √[ f (x)] − 1, 2, , −, , , for | f (x)| > 1, , f ' (x), f (x) √[ f (x)]2 − 1, , , for | f (x)| > 1, , Table 1.2.2, Some Important Formulae for Inverse Trigonometric Functions :, (1) sin−1 (sin θ) = θ, sin(sin−1 x) = x, (3) tan−1 (tan θ) = θ, tan(tan−1 x) = x, , (2) cos−1 (cos θ) = θ, cos(cos−1 x) = x, (4) cot−1 (cot θ) = θ, cot(cot−1 x) = x, , (5) sec−1 (sec θ) = θ, sec(sec−1 x) = x, , (6) cosec−1 (cosec θ) = θ, cosec(cosec−1 x) = x, , (7) sin−1 (cos θ) = sin−1 sin, , π, −θ, 2, , =, , π, −θ, 2, , (8) cos−1 (sin θ) = cos−1 cos, , π, −θ, 2, , =, , π, −θ, 2, , (9) tan−1 (cot θ) = tan−1 tan, , π, −θ, 2, , =, , π, −θ, 2, , (10) cot−1 (tan θ) = cot−1 cot, , π, −θ, 2, , =, , π, −θ, 2, , (11) sec−1 (cosec θ) = sec−1 sec, , π, −θ, 2, , (12) cosec−1 (sec θ) = cosec−1 cosec, 1, x, , (13) sin−1 (x) = cosec−1, , =, , π, −θ, 2, , π, −θ, 2, =, , π, −θ, 2, 1, x, , (14) cosec−1 (x) = sin−1, , (15) cos−1 (x) = sec−1, , 1, x, , (16) sec−1 (x) = cos−1, , 1, x, , (17) tan−1 (x) = cot−1, , 1, x, , (18) cot−1 (x) = tan−1, , 1, x, , (19) sin−1 x + cos−1 x =, , π, 2, , (22) tan−1 x + tan−1 y = tan−1, , (20) tan−1 x + cot−1 x =, x+y, 1 − xy, , π, 2, , (21) sec−1 x + cosec−1 x =, , (23) tan−1 x − tan−1 y = tan−1, , , In, above tables, x is a real variable with restrictions. Table 1.2.3, , 20, , x−y, 1 + xy, , π, 2
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Some Important Substitutions :, Substitutions, , Expression, , √1 − x2, , x = sin θ or x = cos θ, , 2x, , √1 − x2, , x = tan θ or x = cot θ, , √ x2 + 1, , x = sec θ or x = cosec θ, , Expression, , 1 + x2, 1 − x2, , a+x, or, a−x, , a−x, a+x, , x = a cos 2θ or x = a cos θ, , 1+x, or, 1−x, , 1−x, 1+x, , x = cos 2θ or x = cos θ, , a + x2, or, a − x2, , a − x2, a + x2, , 1 + x2, 3x − 4x3 or 1 − 2x2, 4x3 − 3x or 2x2 − 1, 3x − x3, 1 − 3x2, 1 − [ f (x)]2, 2 f (x), or, 1 + [ f (x)]2 1 + [ f (x)]2, , x2 = a cos 2θ or x2 = a cos θ, , Substitutions, x = tan θ, x = tan θ, x = sin θ, x = cos θ, x = tan θ, f (x) = tan θ, , Table 1.2.4, SOLVED EXAMPLES, Ex. 1 : Using derivative prove that sin−1 x + cos−1 x =, Solution :, , Let f (x) = sin−1 x + cos−1 x, , . . . . . (I), π, , We have to prove that f (x) =, 2, Differentiate (I) w. r. t. x, d, d, [ f (x)] = [sin−1 x + cos−1 x], dx, dx, 1, 1, f ' (x) =, −, =0, √1 − x2 √1 − x2, , π, ., 2, , f ' (x) = 0 ⇒ f (x) is a constant function., , Let f (x) = c. For any value of x, f (x) must be c only. So conveniently we can choose x = 0,, ∴ from (I) we get,, π π, π, π, f (0) = sin−1 (0) + cos−1 (0) = 0 + = ⇒ c = ∴ f (x) =, 2 2, 2, 2, π, Hence, sin−1 x + cos−1 x = ., 2, Ex. 2 : Differentiate the following w. r. t. x., (i) sin−1 (x3), , , (iv) cot−1, , 1, x2, , (ii) cos−1 (2x2 − x) (iii) sin−1 (2x), (v) cos−1, , 21, , 1+x, 2, , (vi), , sin2 (sin−1 (x2))
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y = tan−1 (3x + 2) + tan−1 (2x − 1), Differentiate w. r. t. x., dy, d, =, [tan−1 (3x + 2) + tan−1 (2x − 1)], dx, dx, d, d, =, [tan−1 (3x + 2)] + [tan−1 (2x − 1)], dx, dx, 1, 1, d, d, =, (3x + 2) +, (2x − 1), 2 ·, 2 ·, 1 + (3x + 2) dx, 1 + (2x − 1) dx, 3, 2, dy, ∴, =, 2 +, 1 + (3x + 2), 1 + (2x − 1)2, dx, EXERCISE 1.2, (6) Differentiate the following w. r. t. x., (i) tan−1 (log x), (ii) cosec−1 (e−x ), , (1) Find the derivative of the function y = f (x), using the derivative of the inverse function, x = f −1 ( y) in the following, (i) y = √x, (iii), (v) y = 2x + 3, , (iii) cot−1 (x3), , (ii), (iv), (vi), , y = 2 − √x, y = log (2x − 1), y = ex − 3, x, (vii) y = e2x − 3, (viii) y = log2, 2, (2) Find the derivative of the inverse function of, the following, (i) y = x2·ex, (ii) y = x cos x, x, (iii) y = x·7, (iv) y = x2 + log x, (v) y = x log x, (3) Find the derivative of the inverse of the, following functions, and also find their value, at the points indicated against them., (i), (ii), , y = x5 + 2x3 + 3x,, y = ex + 3x + 2,, , (v), , π, 2, , tan−1 x + cot−1 x =, , (ii), , sec−1 x + cosec−1 x =, , (vi), , sin−1, , sin4 [sin−1 (√x)], , (7) Differentiate the following w. r. t. x., (i), , cot−1 [cot (ex )], , (ii), , cosec−1, , 2, , 1 − cos (x2), , (iv) cos−1, (v) tan, , −1, , 2, , 1 − tan ( 2x ), 1 + tan ( 2x ), , (vi) cosec−1, (vii) tan, , −1, , π, . . . [for | x | ≥ 1], 2, , (viii) cot−1, , 29, , 1, cos (5x), 1 + cos x, 2, , (iii) cos−1, , 1 + x2, 2, 3, , (ix) cos3 [cos−1 (x3)] (x), , at x = 1, , (i), , cot−1 (4x ), , (viii) sin−1 (x 2 ), , (vii) cos−1 (1 − x2), , at x = 0, 2, 3, (iii) y = 3x + 2 log x , at x = 1, (iv) y = sin (x − 2) + x2, at x = 2, (4) If f (x) = x3 + x − 2, find ( f −1)' (0)., (5) Using derivative prove, , tan−1 (√x), , (iv), , 1, 4 cos3 2x − 3 cos 2x, , 1 + cos ( 3x ), sin ( 3x ), sin 3x, 1 + cos 3x
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cos 7x, 1 + sin 7x, , (ix) tan−1, , 1 + cos x, 1 − cos x, , (x) tan, , −1, , (v), , (xi) tan−1 (cosec x + cot x), (xii) cot, , 1 + sin ( 4x3 ) − 1 − sin ( 4x3 ), , (ii), , cos, , (v), , cos−1, , √3 cos x − sin x, 2, , 3 cos 3x − 4 sin 3x, 5, 3 cos (ex) + 2 sin (ex), , √13, , (vi) cosec−1, , 10, 6 sin (2 ) − 8 cos (2x), x, , (9) Differentiate the following w. r. t. x., (i), , cos−1, , 1 − x2, 1 + x2, , (ii) tan−1, , 1 + 9x, , (ix) sin−1, , 1 − 25x2, 1 + 25x2, , (xi) tan, , 2x 2, 1 − x5, , ex − e −x, ex + e −x, 1, , (viii) sin, , 4x + 2, 1 + 24x, , (x) sin−1, , 1 − x3, 1 + x3, , (xii) cot−1, , 1 − √x, 1 + √x, , −1, , (10) Differentiate the following w. r. t. x., , √2, , (iv) cos−1, , 1 − 9x, , −1, , −1, , cos √x + sin √x, , (iii) sin−1, , (iv) sin−1 (2x √1 − x2), , 5, , (8) Differentiate the following w. r. t. x., 4 sin x + 5 cos x, (i) sin−1, √41, −1, , 1 + x2, , cos−1 (3x − 4x3) (vi) cos−1, , (vii) cos, , 1 + sin ( 4x3 ) + 1 − sin ( 4x3 ), , −1, , 1 − x2, , (iii) sin−1, , (i) tan−1, , 8x, 1 − 15x2, , (ii) cot−1, , 1 + 35x2, 2x, , (iii) tan−1, , 2√x, 1 + 3x, , (iv) tan−1, , 2x + 2, 1 − 3(4x), , (v) tan−1, , 2x, 1 + 22x + 1, , (vi) cot−1, , a2 − 6x2, 5ax, , (vii) tan−1, , a + b tan x, b − a tan x, , (viii) tan−1, , 5− x, 6x − 5x − 3, , (ix) cot−1, , 2x, 1 − x2, , 2, , 4 − x − 2x2, 3x + 2, , 1.3.1 Logarithmic Differentiation, The complicated functions given by formulas that involve products, quotients and powers can often, be simplified more quickly by taking the natural logarithms on both the sides. This enables us to use, the laws of logarithms to simplify the functions and differentiate easily. Especially when the functions, are of the form y = [ f (x)]g(x) it is recommended to take logarithms on both the sides which simplifies to, log y = g(x). log [ f (x)], now it becomes convenient to find the derivative. This process of finding the, derivative is called logarithmic differentiation., SOLVED EXAMPLES, Ex. 1 : Differentiate the following w. r. t. x., (i), , (x2 + 3)2, , 3, , (x3 + 5)2, , (2x2 + 1)3, , 2, , , , (ii), , 30, , x, , ex (tan x) 2, 3, , (1 + x2) 2 cos3 x
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3, , 5, , 2, , (iii) (x + 1) 2 (2x + 3) 2 (3x + 4) 3 for x ≥ 0, , (iv), , xa + xx + ax, , (v) (sin x)tan x − xlog x, , Solution :, (x2 + 3)2, , (i) Let y =, , (x3 + 5)2, , 3, , (2x2 + 1)3, , Taking log of both the sides we get,, (x2 + 3)2, , log y = log, , 3, , 2, , (x3 + 5)2, , (x2 + 3)2 (x3 + 5) 3, , = log, , (2x2 + 1)3, , 3, , (2x2 + 1) 2, , 2, , 3, , , = log (x2 + 3)2 (x3 + 5) 3 − log (2x2 + 1) 2, 2, , 3, , , = log (x2 + 3)2 + log (x3 + 5) 3 − log (2x2 + 1) 2, log y = 2 log (x2 + 3) +, Differentiate w. r. t. x., , 2, 3, log (x3 + 5) − log (2x2 + 1), 3, 2, , 2, 3, d, d, (log y) =, 2 log (x2 + 3) + log (x3 + 5) − log (2x2 + 1), 3, 2, dx, dx, d, 2 d, 3 d, 1 dy, , = 2· [log (x2 + 3)] + · [log (x3 + 5)] − · [log (2x2 + 1)], dx, 3 dx, 2 dx, y dx, 2, 3, 2, d, d, d, · (x3 + 5) −, · (2x2 + 1), = 2 · (x2 + 3) + 3, 2, 3(x + 5) dx, 2(2x + 1) dx, x + 3 dx, 2, 3, 2, dy, 2, = y 2, (2x) +, (3x, ), −, (4x), 3(x3 + 5), 2(2x2 + 1), x +3, dx, (x2 + 3)2 3 (x3 + 5), dy, ∴ , =, dx, (2x2 + 1)3, , 2, , 2, , (ii) Let y =, , 2x2, 6x, 4x, +, − 2, 3, 2, x + 1 (x + 5) 2x + 1, , x, , ex (tan x) 2, 3, , (1 + x2) 2 cos3 x, , Taking log of both the sides we get,, 2, , log y = log, , 3, 2 2, , (1 + x ) (cos x)3, , , , , , x, , ex (tan x) 2, , ∴, , 3, , x, , 2, , = log ex (tan x) 2 − log (1 + x2) 2 (cos x)3, x, , 2, , 3, , = log ex + log (tan x) 2 − log (1 + x2) 2 + log (cos x)3, 3, x, log (tan x) − log (1 + x2) − 3 log (cos x), 2, 2, 3, x, 2, log y = x + log (tan x) − log (1 + x2) − 3 log (cos x), 2, 2, = x2 log e +, , 31
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Differentiate w. r. t. x., 3, d, d 2 x, (log y) =, x + log (tan x) − log (1 + x2) − 3 log (cos x), 2, 2, dx, dx, d, d x, 3 d, x d, 1 dy, d 2, −, [log (1 + x2)] − 3 [log (cos x)], =, (x ) + · [log (tan x)] + log (tan x), dx, dx 2, 2 dx, 2 dx, y dx dx, 3, d, 1, 3 d, x 1 d, · (tan x) + log (tan x) −, · (1 + x2) −, · (cos x), = 2x + ·, 2, 2 2(1 + x ) dx, cos x dx, 2 tan x dx, 3, 1, 3, x, = 2x + ·(cot x) (sec2 x) + log (tan x) −, ·(2x), −, (− sin x), 2(1 + x2), 2, cos x, 2, 3x, 1, 1, x cos x, ×, + log (tan x) −, + 3 tan x, =, 2x + ×, 2, 1 + x2, 2 sin x cos x 2, 3x, 1, x, dy, + log (tan x) −, + 3 tan x, = y 2x +, 1 + x2, 2 sin x cos x 2, dx, 2, , x, , 3x, ex (tan x) 2, 1, dy, ∴ , =, 2x + x cosec 2x + log (tan x) −, + 3 tan x, 3, 1 + x2, 2, dx (1 + x2) 2 cos3 x, 3, , 5, , 2, , (iii) Let y = (x + 1) 2 (2x + 3) 2 (3x + 4) 3, Taking log of both the sides we get,, 3, , 5, , 2, , log y = log (x + 1) 2 (2x + 3) 2 (3x + 4) 3, 3, , 5, , 2, , , = log (x + 1) 2 + log (2x + 3) 2 + log (3x + 4) 3, 3, 5, 2, log (x + 1) + log (2x + 3) + log (3x + 4), 2, 2, 3, Differentiate w. r. t. x., log y =, , 5, 2, d, d 3, (log y) =, log (x + 1) + log (2x + 3) + log (3x + 4), 2, 3, dx, dx 2, 1 dy, 3 d, 5 d, 2 d, ·, = ·, [log (2x + 1)] + · [log (3x + 2)] +, · [log (3x + 4)], y dx, 2 dx, 2 dx, 3 dx, 3, d, 5, d, 2, d, ·, (2x + 1) +, ·, (3x + 2) +, ·, (3x + 4), =, 2(2x + 1) dx, 2(3x + 1) dx, 3(3x + 4) dx, 5, 2, 3, dy, (2) +, (3) +, (3), = y, 2(3x + 1), 3(3x + 4), 2(2x + 1), dx, 3, 5, 2, 15, 2, 3, dy, +, +, ∴ , = (x + 1) 2 (2x + 3) 2 (3x + 4) 3, 2x + 1 2(3x + 1) 3x + 4, dx, (iv) Let y = xa + xx + ax, Here the derivatives of xa and ax can be found directly but we can not find the derivative of xx, without the use of logarithm. So the given function is split in to two functions, find their derivatives, and then add them., , 32
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Let u = xa + ax and v = xx, ∴ y = u + v, , where u and v are differentiable functions of x., dy du dv, =, + , . . . . . (I), dx dx dx, Now, u = xa + ax, Differentiate w. r. t. x., du d a, d, =, (x ) + (ax), dx dx, dx, du, = axa −1 + ax log a , . . . . . (II), dx, And, v = xx, Taking log of both the sides we get,, log v = log xx, log v = x log x, Differentiate w. r. t. x., d, d, (log v) =, (x log x), dx, dx, 1 dv, d, d, , = x, (log x) + log x (x), v dx, dx, dx, 1, dv, = v x × + log x (1), x, dx, dv, = xx [1 + log x] . . . . . (III), dx, Substituting (II) and (III) in (I) we get,, dy, = axa−1 + ax log a + xx [1 + log x], dx, Let y = (sin x)tan x − xlog x, Let u = (sin x)tan x and v = xlog x, ∴ y = u − v, , where u and v are differentiable functions of x., dy du dv, =, − , . . . . . (I), dx dx dx, Now, u = (sin x)tan x , taking log of both the sides we get,, , (v) , , log u = log (sin x)tan x, ∴, log u = tan x log (sin x), Differentiate w. r. t. x., d, d, (log u) =, [tan x log (sin x)], dx, dx, 1 du, d, d, , = tan x, [log (sin x)] + log (sin x) (tan x), u dx, dx, dx, 1 d, · ·(sin x) + log (sin x)·(sec2), = tan x·, sin x dx, , 33
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du, 1, = u tan x·, ·(cos x) + sec2 x·log (sin x), dx, sin x, du, = (sin x)tan x [tan x·cot x + sec2 x·log (sin x)], dx, du, . . . . . (II), = (sin x)tan x [1 + sec2 x·log (sin x)], dx, And, v = xlog x, Taking log on both the sides we get,, log v = log (xlog x), log v = log x log x = (log x)2, Differentiate w. r. t. x., d, d, (log v) =, [(log x)2], dx, dx, d, 1 dv, = 2 log x, (log x), , dx, v dx, dv, 2 log x, 2xlog x log x, , =u, =, , dx, x, x, , . . . . . (III), , Substituting (II) and (III) in (I) we get,, 2xlog x log x, dy, , = (sin x)tan x [1 + sec2 x·log (sin x)] −, dx, x, 1.3.2 Implicit Functions, Functions can be represented in a variety of ways. Most of the functions we have dealt with so far, have been described by an equation of the form y = f (x) that expresses y solely in terms of the variable, x. It is not always possible to solve for one variable explicitly in terms of another. Those cases where it, is possible to solve for one variable in terms of another to obtain y = f (x) or x = g (y) are said to be in, explicit form., If an equation in x and y is given but x is not an explicit function of y and y is not an explicit function, of x then either of the variables is an Implicit function of the other., 1.3.3 Derivatives of Implicit Functions, 1., , Differentiate both sides of the equation with respect to x (independent variable), treating y as a, differentiable function of x., , 2., , Collect the terms containing, , dy, dy, on one side of the equation and solve for ., dx, dx, , 34
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(v) Differentiate 3x w. r. t. logx 3., , (4) (i) Differentiate x sin x w. r. t. tan x., 2x, (ii) Differentiate sin−1, 1 + x2, w. r. t. cos−1, , 1 − x2, ., 1 + x2, , (iii) Differentiate tan−1, w. r. t. sec−1, , x, , (vii) Differentiate xx w. r. t. xsin x., (viii) Differentiate tan−1, , ., , 2x2 − 1, , cos x, 1 + sin x, , w. r. t. sec−1 x., , √ 1 − x2, , 1, , (iv) Differentiate cos−1, , (vi) Differentiate tan−1, , 1 − x2, 1 + x2, , , , w. r. t. tan−1 x., , w. r. t. tan−1, , √ 1 + x2 − 1, x, , 2x√ 1 − x2, ., 1 − 2x2, , 1.5.1 Higher order derivatives :, If f (x) is differentiable function of x on an open interval I, then its derivative f ' (x) is also a function, on I, so f ' (x) may have a derivative of its own, denoted as ( f ' (x))' = f '' (x). This new function f '' (x) is, called the second derivative of f (x). By Leibniz notation, we write the second detivative of, d dy, d2 y, = 2, dx dx, dx, By method of first principle, y = f (x) as y'' = f '' (x) =, , lim, f ' (x) = h→0, , f (x + h) − f (x), dy, =, and, h, dx, , f '' (x) = lim, h→0, , f ' (x + h) − f ' (x), d2 y, = 2, h, dx, , d, [ f '' (x)] = f ''' (x)., dx, Now the new function f ''' (x) is called the third derivative of f (x). We write the third of y = f (x) as, Further if f '' (x) is a differentiable function of x then its derivative is denoted as, , y ''' = f ''' (x) =, f (4) (x) =, , d4 y, dx4, , d d2 y, d3 y, =, . The fourth derivative, is usually denoted by f (4) (x). Therefore, dx dx2, dx3, , ., , In general, the nth derivative of f (x), is denoted by f (n) (x) and it obtained by differentiating f (x),, dn y, n times. So, we can write the nth derivative of y = f (x) as y(n) = f (n) (x) = n . These are called higher order, dx, derivatives., Note : The higher order derivatives may also be denoted by y2, y3, . . . , yn., , 49
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dy, dy d2 y, dy 2, · 2 − 2x, = 2m2y, dx, dx dx, dx, dy, Dividing throughout by 2 we get,, dx, 2, dy, dy, , ( 1 − x2)· 2 − x, = m2y, dx, dx, 2( 1 − x2)·, , , , , , ∴ ( 1 − x2)·, , d2 y, dx, , 2, , −x, , dy, dx, , − m2y = 0, , 1.5.2 Successive differentiation (or nth order derivative) of some standard functions :, Successive Differentiation is the process of differentiating a given function successively for n times, and the results of such differentiation are called successive derivatives. The higher order derivatives are, of utmost importance in scientific and engineering applications., There is no general formula to find nth derivative of a function. Because each and every function has, it's own specific general formula for it's nth derivative. But there are algorithms to find it., So, here is the algorithm, for some standard functions., Let us use the method of mathematical induction whereever applicable., Step 1 :- Use simple differentiation to get 1st, 2nd and 3rd order derivatives., Step 2 :- Observe the changes in the coefficients, the angles, the power of the function and the signs of, each term etc., Step 3 :- Express the nth derivative with the help of the patterns of changes that you have observed., This will be your general formula for the nth derivative of the given standard function., SOLVED EXAMPLES, Ex. 1 : Find the nth derivative of the following :, (i) xm , , (ii), , , , (v), , (iv) sin x , Solution :, (i) Let y = xm, , 1 , ax + b, cos (ax + b) , , (iii) log x, (vi) eax sin (bx + c), , Differentiate w. r. t. x, d d2 y, d m−2, = m·(m − 1), (x ), 2, dx dx, dx, d3 y, = m·(m − 1)·(m − 2) xm − 3, dx3, In general nth order derivative will be, dn y, = m·(m − 1)·(m − 2)...[m − (n − 1)] xm − n, dxn, dn y, = m·(m − 1)·(m − 2)...[m − n + 1] xm − n, dxn, , Differentiate w. r. t. x, dy d m, =, (x ) = mxm − 1, dx dx, Differentiate w. r. t. x, d m−1, d dy, =m, x, dx, dx dx, d2 y, 2 = m·(m − 1) xm − 2, dx, , 56
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case (i) :-, , If m > 0 and m > n, then, dn y m·(m − 1)·(m − 2)... [m − (n − 1)] · (m − n)... 2·1 m − n, =, x, dxn, (m − n) · [m − n − 1]... 2·1, dn y m!. xm − n, =, dxn (m − n)!, , case (ii) :-, , If m > 0 and m = n, then, dn y n!. xm − n, n!. x0, =, =, = n!, dxn (n − n)!, 0!, , (ii) Let y =, , case (iii) :-, , 1, ax + b, , If m > 0 and m < n, then, dn y, =0, dxn, , (iii) Let y = log x, Differentiate w. r. t. x, dy d, 1, =, (log x) =, dx dx, x, , Differentiate w. r. t. x, dy d, d, 1, −1, =, =, ·, (ax + b), dx dx ax + b (ax + b)2 dx, , Differentiate w. r. t. x, , dy (− 1)·a, =, dx (ax + b)2, , d 1, d dy, =, dx x, dx dx, , Differentiate w. r. t. x, , d2 y, , =, , − 1 (− 1)1, = 2, x2, x, , d, d dy, 1, = (− 1)(a), , dx (ax + b)2, dx dx, , Differentiate w. r. t. x, , dy, d, −2, 2 = (− 1)(a), ·, (ax + b), dx, (ax + b)3 dx, , d d2 y, d 1, = (− 1)1, 2, dx dx, dx x2, , d2 y (− 1)2·2·1·a2, 2 =, dx, (ax + b)3, , d3 y, , dx2, , 2, , dx3, , Differentiate w. r. t. x, d, d dy, 1, = (− 1)2·2·1·a2·, , 2, dx (ax + b)3, dx dx, , dn y (− 1)n − 1·1·2·3... (n − 1), =, dxn, xn, , d3 y, d, −3, 3 = (− 1)2·2·1·a2·, · (ax + b), 4, dx, (ax + b) dx, , dn y, dxn, , d y (− 1) ·3·2·1·a, 3 =, dx, (ax + b)4, , , 3, , (− 1)2·1·2, −2, =, x3, x3, , In general nth order derivative will be, , 2, , 3, , = (− 1)1, , 3, , In general nth order derivative will be, , dn y (− 1)n·n·(n − 1)... 2·1·an, n =, dx, (ax + b)n + 1, dn y (− 1)n·n!·an, n =, dx, (ax + b)n + 1, , 57, , =, , (− 1)n − 1·(n − 1)!, xn
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Let us Remember, ֍, , If a function f (x) is differentiable at x = a then it is continuous at x = a, but the converse is not, true., , Chain Rule : If y is differerentiable function of u and u is differerentiable function of x then y, dy dy du, is differerentiable function of x and, =, ·, dx du dx, ֍ If y = f (x) is a differentiable of x such that the inverse function x = f −1( y) exists then, 1, dy, dy, ,, where, ≠0, =, , dy, dx, dx, dx, ֍, , ֍, , Derivatives of Inverse Trigonometric functions :, f (x), , sin −1 x, , f ' (x), , 1, ,, √1 − x2, |x| < 1, , ֍, , −, , cos −1 x, , tan −1 x, , cot −1 x, , sec −1 x, , 1, ,, √1 − x2, |x| < 1, , 1, 1 + x2, , −, , 1, 1 + x2, x∈R, , 1, x√ x2 − 1, |x| < 1, , x∈R, , cosec −1 x, 1, x√ x2 − 1, |x| < 1, , −, , This is a simple shortcut to find the derivative of (function) (function), d g, f =f, dx, , g, , g, · f ' + (log f ) · g', f, , If y = f (t ) and y = g (t ) is a differentiable of t such that y is a function of x then, dy, dy, dx, dt, , ,, where, ≠0, =, dx, dx, dt, dt, ֍ Implict function of the form x m y n = (x + y) m + n , m, n ∈ R always have the first order derivative , dy y, d2y, =, and second order derivative 2 = 0, dx x, dx, , ֍, , MISCELLANEOUS EXERCISE 1, (I) Choose the correct option from the given alternatives :, 1, 8, (1) Let f (1) = 3, f ' (1) = − , g (1) = − 4 and g' (1) = − . The derivative of √ [ f (x)]2 + [g (x)]2, 3, 3, , w. r. t. x at x = 1 is, 29, 7, 31, 29, (A) −, (B), (C), (D), 15, 3, 15, 15, , 61
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(2), , If y = sec (tan−1 x) then, , (A), (3), , 1+ 24x, , (B), , 1 + 42x, , dy, = ..., dx, x ( x log y − y ), , y ( y log x − x ), , 2 − 4x2, , √ 1 − x2, x, 1 + √ 1 − x2, , x, , (D) √ 2, , √2, , 4x + 1 log 2, 1 + 42x, , (C), , 4x + 1 log 4, 1 + 44x, , (D), , 22(x + 1) log 2, 1 + 24x, , y ( y log x − x ), , (C), , x ( x log y − y ), , dy, = ..., dx, 2 + 4x2, (B), √ 1 − x2, , (C), , y2 (1 − log x ), x2 (1 − log y ), , 4x2 − 1, , 1−x, dy, , then, = ..., 1+x, dx, , 1 − 2x, , 1 − 2x, , (C), , √ 1 − x2, , (D), , √ 1 − x2, , + sin 2 tan−1, , (D), , (D), , 2 √ 1 − x2, , y (1 − log x ), x (1 − log y ), , 1 − 2x2, , √ 1 − x2, , 1 − 2x2, , √ 1 − x2, , If y is a function of x and log (x + y) = 2xy, then the value of y' (0) = ..., (B) 0, , (C) −1, , If g is the inverse of a function f and f ' (x) =, (B), , (A) 1 + x7, (9), , (B), , (B), , √ 1 − x2, , (A) 2, (8), , 1, , , which of the following is not the derivative of f (x), , 2·4x log 4, , If y = tan−1, , (A), (7), , (C), , 1, , 4x + 2, , If y = sin (2 sin−1 x), then, , (A), (6), , at x = 1, is equal to :, , If x y = y x, then, , (A), (5), , dx, , (B) 1, , If f (x) = sin−1, , (A), (4), , 1, 2, , dy, , 1, 7, 1 + [g(x)], , If x √ y + 1 + y √ x + 1 = 0 and x ≠ y then, , (A), , 1, (1 + x)2, , (10) If y = tan−1, (A), , x, , √ a2 − x2, , (B) −, a−x, a+x, , 1, (1 + x)2, , 1, , then the value of g' (x) is equal to :, 1 + x7, 7, , dy, dx, , a, , (C) 1 + [g(x)], , (D) 7x6, , (C) (1 + x)2, , (D) −, , = ..., , , where − a < x < a then, (B), , (D) 1, , dy, dx, , = ..., (C) −, , √ a2 − x2, , 62, , x, x+1, , 1, 2√ a2 − x2, , (D), , 1, 2√ a2 − x2
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(11) If x = a (cos θ + θ sin θ), y = a (sin θ − θ cos θ) then, (A), , 8 √2, , (B) −, , aπ, , (12) If y = a cos (log x) and A, (A) x2, −x, −y, , 8 √2, , d2 y, dx2, , π, 4, , = ..., , aπ, , (C), , aπ, , θ=, , 8 √2, , (D), , 4 √2, aπ, , dy, + C = 0, then the values of A, B, C are ..., dx, dx2, (B) x2, x, y, (C) x2, x, −y, (D) x2, −x, y, , d2 y, , +B, , (II) Solve the following :, (1), , f (x) = −x,, , for −2 ≤ x < 0, , g(x) = 6 − 3x,, 2x − 4, =, ,, 3, , = 2x,, for 0 ≤ x ≤ 2, 18 − x, =, ,, for 2 < x ≤ 7, 4, Let u(x) = f [ g(x)], v(x) = g [ f (x)] and w(x) = g [ g(x)]., , (2), , for 0 ≤ x ≤ 2, for 2 < x ≤ 7, , Find each derivative at x = 1, if it exists i.e. find u' (1), v' (1) and w' (1). if it doesn't exist then, explain why ?, The values of f (x), g(x), f ' (x) and g' (x) are given in the following table., f (x), 3, 2, , x, −1, 2, , g(x), 2, −1, , f ' (x), −3, −5, , g' (x), 4, −4, , Match the following., A Group - Function, d, [ f (g (x))] at x = −1, dx, d, [ g ( f (x) − 1)] at x = −1, (B), dx, d, [ f ( f (x) − 3)] at x = 2, (C), dx, d, [ g ( g (x))] at x = 2, (D), dx, (A), , (3), , B Group - Derivative, 1. −16, 2., , 20, , 3. −20, 4., , 15, , 5., , 12, , Suppose that the functions f and g and their derivatives with respect to x have the following, values at x = 0 and x = 1., x, , f (x), , g(x), , 0, , 1, , 1, , 1, , 3, , −4, , f ' (x) g' (x), 5, −, , 1, 3, , 1, 3, 8, −, 3, , (i) The derivative of f [ g(x)] w. r. t. x at x = 0 is ......, (ii) The derivative of g [ f (x)] w. r. t. x at x = 0 is ......, d 10, (iii) The value of, [x + f (x)]−2, is ......, x=1, dx, (iv) The derivative of f [(x + g(x)] w. r. t. x at x = 0 is ......, , 63
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(4), , Differentiate the following w. r. t. x, , (i) sin 2 tan−1, , 1−x, 1+x, , , , (iii) tan−1, , √ x (3 − x), , 1 − 3x, , (iv) cos−1, , (v) tan−1, , x, 1 − 10x2, −1, +, cot, 1 + 6x2, 7x, , (vi) tan−1, , (5), , 1+x, 1−x, , (ii) sin2 cot−1, , y2, , dy y, (i) If √ y + x + √ y − x = c, then show that, = −, dx x, , x2, , √1 + x − √1 − x, 2, , √ 1 + x2 + x, √ 1 + x2 − x, , − 1., , 1 − y2, dy, (ii) If x √ 1 − y2 + y √ 1 − x2 = 1, then show that, =−, ., 1 − x2, dx, dy sin2 (a + y), (iii) If x sin (a + y) + sin a cos (a + y) = 0, then show that, =, ., sin a, dx, (iv) If sin y = x sin (a + y), then show that, x, , (v) If x = e y , then show that, , dy sin2 (a + y), =, ., sin a, dx, , dy, x−y, =, ., dx x log x, , (vi) If y = f (x) is a differentiable function then show that, (6), , (i) Differentiate tan−1, , , , (ii) Differentiate log, , , , (iii) Differentiate tan−1, , d2 x, dy2, , =−, , dy, dx, , √ 1 + x2 − 1 w. r. t. tan−1 2x √ 1 − x2 ., 2, 1 − 2x, , x, , √ 1 + x2 + x, w. r. t. cos (log x)., √ 1 + x2 − x, √ 1 + x2 − 1, , w. r. t. cos−1, , 1 + √ 1 + x2, , ., 2√ 1 + x2, d2 y a2 b2, 2, 2, 2, 2, 2, (7) (i) If y = a cos x + b sin x , show that y + 2 = 3 ., dx, y, 2, d, dy, y, + (4x2 + 3) y = 0., (ii) If log y = log (sin x) − x2 , show that 2 + 4x, dx, dx, d2 y, dy 2, (iii) If x = a cos θ, y = b sin θ, show that a2 y 2 +, + b2 = 0., dx, dx, x, , (iv) If y = A cos (log x) + B sin (log x), show that x2 y2 + x y1 + y = 0., (v) If y = A emx + B enx, show that y2 − (m + n) y1 + (mn) y = 0., , v, , v, 64, , v, , −3, , ·, , d2 y, ., dx2
